Simplifying Expressions

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SAT Subject Test in Math I › Simplifying Expressions

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Simplify the expression.

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}$

$\small \frac{xy^3}{z^3}$

CORRECT

$\small \frac{x^2y^5}{z^3}$

$\small \frac{z^3}{x^2y^5}$

$\small \frac{z^3}{xy^3}$

Explanation

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}$

Because we are only multiplying terms in the numerator, we can disregard the parentheses.

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}=\frac{x^2y^2x^3y^4z}{x^4y^3z^4}$

To combine like terms in the numerator, we add their exponents.

$\small \frac{x^5y^6z}{x^4y^3z^4}$

To combine like terms between the numerator and denominator, subtract the denominator exponent from the numerator exponent.

$x^{5-4}y^{6-3}z^{1-4}=xy^3z^{-3}$

Remember that any negative exponents stay in the denominator.

$\frac{xy^3}{z^3}$

Give the value of that makes the polynomial $9x^{2}+ Nx + 100$ the square of a linear binomial.

CORRECT

None of the other responses gives a correct answer.

Explanation

A quadratic trinomial is a perfect square if and only if takes the form

$A^{2} \pm 2AB + B^{2}$ for some values of and .

$9x^{2}+ Nx + 100 = (3x)^{2} + Nx + 10^{2}$ , so

and .

For $9x^{2}+ Nx + 100$ to be a perfect square, it must hold that

$Nx = 2AB = 2 \cdot 3x \cdot 10 = 60x$ ,

so . This is the correct choice.

Simplify the following expression:

CORRECT

Explanation

When simplifying an equation,you must find a common factor for all values in the equation, including both sides.

and, can all be divided by so divide them all at once

$\left ( \frac{3}{3}=1, \frac{18}{3}=6, \frac{81}{3}=27\right )$ .

This leaves you with

Simplify the expression $\frac{x^5(3x^2y^3)}{x^4y^2z}$

$\frac{3x^3y}{z}$

CORRECT

$\frac{3x^6y}{z}$

$\frac{3x^3}{yz}$

Already in simplest form

Explanation

$\frac{x^5(3x^2y^3)}{x^4y^2z}$

Simplify the numerator by multiplying in the $\tiny x^5$ term

$\frac{3x^7y^3}{x^4y^2z}$

Cancel out like terms in the numerator and denominator.

$\frac{3x^3y}{z}$

Simplify:

$\frac{(x^5y^2z^3)(x^2yz)}{x^4y^{-3}z^3}$

CORRECT

Explanation

To simplify, we begin by simplifying the numerator. When muliplying like bases with different exponents, their exponents are added.

For x: $\small 4+3=7$

For y: $\small 2+1=3$

For z: $\small 3+1=4$

The numerator is now .

When dividing like bases, their exponents are subtracted.

For x: $\small 7-4=3$

For y: $\small 3-(-3)=6$

For z: $\small 4-3=1$

Thus, our answer is .

Divide:

$\left ( 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x \right ) \div 8x^{2}$

$\frac{ 3 }{ 2 }x ^{2} + 2x + 8 + \frac{1}{x}$

CORRECT

$\frac{ 3 }{ 2 }x ^{2} + 3x + 8$

$\frac{ 3 }{ 2 }x ^{3} + 2x^{2} + 8x + 1$

$\frac{ 3 }{ 2 }x ^{3} + 3x^{2} + 8x$

$\frac{ 3 }{ 2 }x ^{3} + 2x^{2} + 8x$

Explanation

Divide termwise:

$\left ( 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x \right ) \div 8x^{2}$

$=\frac{ 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x }{ 8x^{2}}$

$=\frac{ 12 x ^{4} }{ 8x^{2}} + \frac{ 16x^{3} }{ 8x^{2}} + \frac{ 64x^{2} }{ 8x^{2}} +\frac{ 8 x }{ 8x^{2}}$

$=\frac{ 12 }{ 8 }x ^{4-2} + \frac{ 16 }{ 8}x^{3-2} + \frac{ 64 }{ 8 } +\frac{ 8 }{ 8} \cdot \frac{1}{x^{2-1}}$

$=\frac{ 3 }{ 2 }x ^{2} + 2x + 8 + \frac{1}{x}$

The polynomial $x^{4} - 27x^{2}+ 5x - A$ is divisible by the linear binomial . Evaluate .

CORRECT

None of the other choices gives the correct answer.

Explanation

By the factor theorem, a polynomial is divisible by the linear binomial if and only if . Therefore, we want the value of that makes the polynomial equal to 0 when evaluated at .

$x^{4} - 27x^{2}+ 5x - A = 0$

$6^{4} - 27 \cdot 6^{2}+ 5 \cdot 6 - A = 0$

$1,296- 27 \cdot 36+ 30- A = 0$

Factor:

$27Q^{3}- 270Q^{2}+900Q -1,000$

$(3Q - 10)^{3}$

CORRECT

$(3Q-10)(9Q^{2}+ 30 Q +100)$

$(3Q-10)(9Q^{2}- 30 Q +100)$

$(3Q-10)(3Q+10)^{2}$

The polynomial is prime.

Explanation

$27Q^{3}- 270Q^{2}+900Q -1,000$

$=\left (3Q \right )^{3}- 3 \cdot 90Q^{2}+3\cdot 300Q -10^{3}$

$=\left (3Q \right )^{3}- 3 \cdot 9Q^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

$=\left (3Q \right )^{3}- 3 \cdot \left (3Q \right )^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

This can be factored out as the cube of a difference, where :

$A^{3}- 3A^{2}B+3AB^{2} -B^{3} = (A - B)^{3}$

Therefore,

$27Q^{3}- 270Q^{2}+900Q -1,000$

$=\left (3Q \right )^{3}- 3 \cdot \left (3Q \right )^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

$= (3Q - 10)^{3}$

Factor completely:

$t^{3} - 243$

The polynomial is prime.

CORRECT

$(t - 3)\left ( t^{2}+9t + 81\right )$

$(t - 3)\left ( t^{2} + 81\right )$

$(t - 3)\left ( t^{2}+ 3t + 81\right )$

$(t - 3)(t-9) ^{2}$

Explanation

Since the first term is a perfect cube, the factoring pattern we are looking to take advantage of is the difference of cubes pattern. However, 243 is not a perfect cube of an integer $(6^{3} = 216 < 243 < 7^{3} = 343)$ , so the factoring pattern cannot be applied. No other pattern fits, so the polynomial is a prime.